Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
class Main
{
    public static void main (String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        double[] p = new double[n], q = new double[n];
        for(int i = 0; i < n; i++){
            p[i] = sc.nextDouble();
            q[i] = 1 - p[i];
        }
        double[][] dp = new double[n][n + 1];
        dp[0][0] = q[0];
        dp[0][1] = p[0];
        for(int i = 1; i < n; i++){
            dp[i][0] = dp[i-1][0] * q[i];
            dp[i][i+1] = dp[i-1][i] * p[i];
            for(int j = 1; j <= i; j++){
                dp[i][j] = dp[i-1][j] * q[i] + dp[i-1][j-1] * p[i];
            }
        }
        double ans = 0;
        for(int i = n / 2 + 1; i <= n; i++){
            ans += dp[n-1][i];
        }
        System.out.println(ans);
    }
}